Critical Points and Integrable Density Singularities

Why a Vanishing Derivative Changes the Density

For a one-to-one differentiable transformation, the density formula divides the input density by the absolute derivative of the transformation. A small derivative therefore stretches probability over a short range of output values and can produce a large density. At a critical point, where the derivative is zero, that formula cannot be evaluated directly. The right question is how quickly the derivative approaches zero and what that implies for the density nearby.

The key distinction is between a density that becomes unbounded near a value and an atom at that value. An unbounded density can still assign probability zero to the value itself; an atom assigns it positive probability. The local shape of the transformed density explains the first phenomenon, while the probability of the entire inverse image of the value determines the second.

Assume \(X\) has density \(f\), and consider a critical point \(x_0\) at which \(f\) is continuous and positive. Suppose the first nonzero derivative of \(g\) at \(x_0\) has finite order \(m\geq 2\). In a neighborhood of \(x_0\), this means \(g(x_0+h)-g(x_0)\) is approximately \(a h^m\), where \(a=g^{(m)}(x_0)/m!\ne 0\). The order \(m\) controls the singularity: higher-order flattening produces a stronger, though still integrable, divergence.

Local Density Near a Critical Value

Away from critical values, the finite-to-one change-of-variables formula sums a contribution from each inverse branch: at a regular value \(y\), the density is the sum of \(f(x)/|g'(x)|\) over roots \(x\) of \(g(x)=y\). Near \(y_0\), focus on the roots close to \(x_0\). Taylor expansion gives \(g(x_0+h)-y_0=a h^m+o(|h|^m)\) and \(g'(x_0+h)=ma h^{m-1}+o(|h|^{m-1})\). Thus a nearby root has distance from \(x_0\) approximately \((|y-y_0|/|a|)^{1/m}\), while the derivative there is of order \(|y-y_0|^{(m-1)/m}\).

Here \(f_{Y,\mathrm{loc}}\) denotes only the contribution from roots near \(x_0\). If other inverse branches also approach regular roots over \(y_0\), they contribute additional terms; they do not change this local leading singularity. The density formula is used at regular values, not by substituting \(y=y_0\) into a denominator that is zero.

Proof. Put \(h=x-x_0\). Taylor’s theorem under the stated differentiability assumptions yields \(g(x_0+h)-y_0=a h^m+o(|h|^m)\) and \(g'(x_0+h)=ma h^{m-1}+o(|h|^{m-1})\) as \(h\to 0\). On each side of \(x_0\), the derivative has a fixed nonzero sign sufficiently close to \(x_0\): its leading term is \(ma h^{m-1}\), and the remainder divided by \(|h|^{m-1}\) tends to zero. The map is therefore strictly monotone on each such side and has a local inverse there.

For a root \(x=x_0+h\) near \(x_0\), the first expansion implies \(|y-y_0|=|a||h|^m(1+o(1))\). Consequently, \(|h|=(|y-y_0|/|a|)^{1/m}(1+o(1))\). The derivative expansion then gives \[ |g'(x)|=m|a|^{1/m}|y-y_0|^{(m-1)/m}(1+o(1)). \] Continuity of \(f\) gives \(f(x)=f(x_0)+o(1)\). Each nearby inverse branch thus contributes \[ \frac{f(x)}{|g'(x)|} \sim \frac{f(x_0)}{m|a|^{1/m}}|y-y_0|^{-(m-1)/m}. \] For even \(m\), \(a h^m\) has the same sign on either side of zero, so there are two local roots on the attainable side and none on the other. For odd \(m\), there is one local root on each side. Summing the branch contributions gives the stated asymptotic.

Finally, \((m-1)/m<1\), and for any \(\varepsilon>0\), \[ \int_0^\varepsilon t^{-(m-1)/m}\,dt=m\varepsilon^{1/m}<\infty. \] The density divergence is therefore integrable. The argument also shows why an infinite limiting density at one point does not require positive probability at that point. \(\square\)

Worked Examples

Critical Values Are Not Automatically Atoms

For any deterministic transformation \(Y=g(X)\), \[ P(Y=y_0)=P\bigl(X\in\{x:g(x)=y_0\}\bigr). \] A zero derivative alone does not imply that this probability is positive. If \(X\) has a density and the level set \(\{x:g(x)=y_0\}\) consists of finitely many points, that set has probability zero, so \(Y\) has no atom at \(y_0\). In the examples above, the critical level sets are finite. By contrast, an atom of \(X\) at a point mapped to \(y_0\) can produce an atom of \(Y\) there, as the mixture example demonstrates. More generally, the level set’s probability—not whether the density is unbounded nearby—decides whether an atom exists.

A common mistake is to substitute the critical root directly into \(f(x)/|g'(x)|\), obtain division by zero, and conclude that the transformed distribution has infinite probability at the critical value. The correction is to use the formula at nearby regular values, determine the rate of divergence, and integrate over a neighborhood. Since the exponent \((m-1)/m\) is less than one, the nearby probability is finite and tends to zero with the neighborhood’s width. Then check \(P(X\in g^{-1}(\{y_0\}))\) separately to decide whether there is an atom.